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Math Help - Logarithm problem

  1. #1
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    Logarithm problem

    state the values of x for which the following identity is true
    log(x+3) + log(x+4) = log(x^2 + 7x + 12)
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  2. #2
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    Hint: Log(xy)=log x + log y and If we have log a, a must be >0
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  3. #3
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by ugie View Post
    state the values of x for which the following identity is true
    log(x+3) + log(x+4) = log(x^2 + 7x + 12)
    This is how I would proceed:

    We rely on the principle that log(xy) = log(x) + log(y).

    LHS means left-hand-side, RHS right-hand-side.

    LHS: log(x+3) + log(x+4) = log((x+3)(x+4)) = log(x^2 + 7x + 12)

    Since this is the same as the RHS, all we need to do is ensure that the quantities (x+3), (x+4) and (x^2 + 7x + 12) are positive -- otherwise, taking their log is undefined.

    So we are solving the simultaneous inequalities

    x + 3 > 0
    x + 4 > 0
    (x+3)(x+4) > 0
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  4. #4
    Senior Member Stroodle's Avatar
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    log(x+3)+log(x+4)=log\left [(x+3)(x+4)\right ]=log(x^2+7x+12)

    So x\in(-3,\infty), as you can't take the log of a negative value.
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